Question: Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{3q^2 + 24q + 36}{-9q^3 + 54q^2 + 144q}$
First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {3(q^2 + 8q + 12)} {-9q(q^2 - 6q - 16)} $ $ x = -\dfrac{3}{9q} \cdot \dfrac{q^2 + 8q + 12}{q^2 - 6q - 16} $ Simplify: $ x = - \dfrac{1}{3q} \cdot \dfrac{q^2 + 8q + 12}{q^2 - 6q - 16}$ Next factor the numerator and denominator. $ x = - \dfrac{1}{3q} \cdot \dfrac{(q + 2)(q + 6)}{(q + 2)(q - 8)}$ Assuming $q \neq -2$ , we can cancel the $q + 2$ $ x = - \dfrac{1}{3q} \cdot \dfrac{q + 6}{q - 8}$ Therefore: $ x = \dfrac{ -q - 6 }{ 3q(q - 8)}$, $q \neq -2$